If 3.0 g N reacts with 4.0 g H, given the equation N2(g) + 3 H2(g) → 2 NH3(g), how many grams of ammonia will form and what is the limiting reactant?

1 Answer
Mar 19, 2018

Answer:

Nitrogen is the limiting reactant, and #3.74# grams of ammonia is formed.

Explanation:

We have the familiar Haber process equation...

#N_2(g)+3H_2(g)rightleftharpoons2NH_3(g)#

Since the mole ratio of #N_2# to #H_2# is #1:3#, then for every mole of nitrogen, we would need three moles of hydrogen.

Nitrogen #(N_2)# has a molar mass of #28 \ "g/mol"#.
Here, if we have #3# grams of nitrogen, i.e.

#(3color(red)cancelcolor(black)"g")/(28color(red)cancelcolor(black)"g""/mol")~~0.11 \ "mol"#

So, we would need #0.11 \ "mol"*3=0.33 \ "mol"# of hydrogen to react with the nitrogen we have.

Hydrogen #(H_2)# has a molar mass of #2 \ "g/mol"#.

But, we have

#(4color(red)cancelcolor(black)"g")/(2color(red)cancelcolor(black)"g""/mol")=2 \ "mol"#

So, hydrogen is the excess reactant and that therefore makes nitrogen the limiting reactant.

That means, we could only produce so much ammonia with only the nitrogen that we have.

The mole ratio between #N_2# and #NH_3# is #1:2#, so one mole of nitrogen gas makes two moles of ammonia gas.

Since we have #0.11# moles of nitrogen, then we make #0.11*2=0.22# moles of ammonia gas.

Ammonia has a molar mass of #17 \ "g/mol"#. So here, we produce

#0.22color(red)cancelcolor(black)"mol"*(17 \ "g")/(color(red)cancelcolor(black)"mol")=3.74 \ "g"#